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# help with series RLC circuit

#### seeeei

Nov 2, 2023
18

determine the characteristic equation for the voltage across the capacitor (iC), the voltage across the inductor (iL) and the voltage across the resistor (iR).

#### seeeei

Nov 2, 2023
18

this is my attempted solution but I'm not sure, also I am not confident in solving the voltage across the capacitor.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,632
VR + VL + VC = 0 V is o.k.
WIth the current direction as shown this directly translates to
$i \times R + L \times \frac{di}{dt} + \frac{q}{C} = 0 V$I don't know why you use negative values in your equation, but this is a mere formal complaint as both your and my equation are equivalent since the result is 0 V and in your 4th line on the left you miraculously come to the same a similar result, but why do you change the integral from $\int i dt$ to $\int v dt$? This is wrong.

Coming to the right side: you make matters unnecessarily complicated by distinguishing IR, IL and IC. Since this circuit has only one loop, IR = IL = IC = I simplifies writing down the equations. You furthermore make it very difficult to follow your writeup when you e.g. $V = L \times \frac{di}{dt}$ without using the subscript "L" for VL.

By the way: It is imho good practice to keep the units into the calculations. This allows to check the result: If the units don't match there is definitely an error in the calculation.

#### seeeei

Nov 2, 2023
18
no that is ∫idt, im sorry if my writing is bad

Last edited:

#### seeeei

Nov 2, 2023
18
VR + VL + VC = 0 V is o.k.
WIth the current direction as shown this directly translates to
$i \times R + L \times \frac{di}{dt} + \frac{q}{C} = 0 V$I don't know why you use negative values in your equation, but this is a mere formal complaint as both your and my equation are equivalent since the result is 0 V and in your 4th line on the left you miraculously come to the same a similar result, but why do you change the integral from $\int i dt$ to $\int v dt$? This is wrong.

Coming to the right side: you make matters unnecessarily complicated by distinguishing IR, IL and IC. Since this circuit has only one loop, IR = IL = IC = I simplifies writing down the equations. You furthermore make it very difficult to follow your writeup when you e.g. $V = L \times \frac{di}{dt}$ without using the subscript "L" for VL.

By the way: It is imho good practice to keep the units into the calculations. This allows to check the result: If the units don't match there is definitely an error in the calculation.

is it okay now?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,632
Looks o.k up to this point:

The next 2 equations are imho not helpful.

The equation now has the form x''+ax'+bx = 0 with a = R/L and b = 1/(LC). This second order differential equation can now be solved using with some help from the internet, see e.g. this page. Once you have the solution, you can re-substitute a and b by the respective expressions.

Btw: Your thread headers are not very informative. Instead of "help meeee I have my answer but I'm not sure" something like "help me solve an LR-circuit" immediately gives our members a hint what your question is about.

#### seeeei

Nov 2, 2023
18
Looks o.k up to this point:
View attachment 61345
The next 2 equations are imho not helpful.

The equation now has the form x''+ax'+bx = 0 with a = R/L and b = 1/(LC). This second order differential equation can now be solved using with some help from the internet, see e.g. this page. Once you have the solution, you can re-substitute a and b by the respective expressions.

Btw: Your thread headers are not very informative. Instead of "help meeee I have my answer but I'm not sure" something like "help me solve an LR-circuit" immediately gives our members a hint what your question is about.
but the question is "determine the characteristic equation for the voltage across the capacitor (iC), the voltage across the inductor (iL) and the voltage across the resistor (iR)." did I just solved the problem asked?

Nov 2, 2023
18

#### seeeei

Nov 2, 2023
18
that is first order, the circuit i have provided is 2nd order circuit

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,632
You sure know how to look up other websites which explain how to solve 2nd order equations. I can't do your work.

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